<u>Answer</u>:
A single atom of an isotope does not have a(n) Neutron.
<u>Explanatio</u>n:
The lack of neutrons are found to be in an isotope of hydrogen. It has one proton and one electron. This protium has an active ingredient pantoprazole. It is found to be an elective proton pump inhibitor, helps in lowering the acid concentration in the stomach. Being used in the treatment of acid problems in stomach and intestine.it can be abundantly found in the oceans. Protium lacks neutron in the nucleus. It is one among the stable isotope. It is also called as ordinary hydrogen.
Answer:
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Explanation:
Approximately 2 mL of Solution A (on the left) is added to a sample of Solution B (on the right) with a dropping pipet. If a precipitate forms, the resulting precipitate is suspended in the mixture. The mixture is then stirred with a glass stirring rod and the precipitate is allowed to settle for about a minute.
Solution A: 0.5 M sodium hydroxide, colorless
Solution B: 0.2 M nickel(II) nitrate, green
Precipitate: light green
Ni(NO3)2(aq) + 2 NaOH(aq) —> Ni(OH)2(s) + 2 NaNO3(aq)
Credits:
Design
Kenneth R. Magnell Central Michigan University, Mt. Pleasant, MI 48859
John W. Moore University of Wisconsin - Madison, Madison, WI 53706
Video
Jerrold J. Jacobsen University of Wisconsin - Madison, Madison, WI 53706
Text
Kenneth R. Magnell Central Michigan University, Mt. Pleasant, MI 48859
Answer:
The initial temperature is 300 K (The temperature doesn't change)
Explanation:
Step 1: Data given
Initial volume = 21L
Final volume = 14L
Initial pressure = 100 kPa = 0.986923 atm
Final pressure = 150 kPa = 1.48038 atm
The final temperature = 300K
Step 2: Calculate the initial temperature
Calculate the initial temperature
(P1*V1)/T1 = (P2*V2)/T2
⇒with P1 = the initial pressure = 0.986923 atm
⇒with V1 = the initial volume = 21 L
⇒ with T1 = the initial temperature = ?
⇒with P2 = the final pressure = 1.48038 atm
⇒with V2 = the final volume = 14 L
⇒with T2 = the final temperature = 300 K
(0.986923 * 21)/T1 = (1.48038*14)/300
T1 = 300 K
The initial temperature is 300 K (The temperature doesn't change)
Answer:
pH = 4.78
Explanation:
<em>There are titrated 65.8mL of a 0.7600M of acetic acid</em>
The reaction of acetic acid, CH₃COOH, with NaOH is:
CH₃COOH + NaOH → CH₃COONa + H₂O
Initial moles of acetic acid are:
0.0658L ₓ (0.7600mol / L) = 0.05000 moles of CH₃COOH
Moles of NaOH added to the acid solution are:
0.0784L ₓ (0.3500mol / L) = 0.02744 moles of NaOH = Moles of CH₃COONa produced -<em>Because NaOH is the limiting reactant-</em>
That means after the reaction moles of CH₃COOH and moles of CH₃COONa are:
CH₃COOH: 0.05000 mol - 0.02744 mol = 0.02256 moles
CH₃COONa: 0.02744 moles
Using H-H equation for acetic acid, pH of the solution is:
pH = 4.70 + log₁₀ [CH₃COONa] / [CH₃COOH]
Replacing:
pH = 4.70 + log₁₀ [0.02744] / [0.02256]
<h3>pH = 4.78</h3>
First write a balanced equation...
P4 + 5 O2 ⇒ 2P2O5
mole of P4 = 136 ÷ (4×31) = 1.09677
MOLE RATION OF P4 : P2O5 IS 1:2
THAT IS 1 mole of p4 reacts to give 2 moleof p2o5
∴ 1.09677 mole............................. (2×1.09677) =2.19....mole
maximum number of mole = 2.19 mol